In the classical consecutive sequential reaction mechanism rigidly defined as , if the second decomp — Chemical Kinetics Chemistry Question
Question
In the classical consecutive sequential reaction mechanism rigidly defined as $A \xrightarrow{k_1} B \xrightarrow{k_2} C$, if the second decomposition step is extraordinarily faster than the first generating step ($k_2 \gg k_1$), what is the expected steady-state mathematical approximation for the concentration of the highly reactive intermediate B?
Answer: A
💡 Solution & Explanation
Because the intermediate B is highly reactive ($k_2 \gg k_1$), we apply the Steady State Approximation (SSA), setting $\frac{d[B]}{dt} \approx 0$. The differential rate equation for B is $\frac{d[B]}{dt} = k_1[A] - k_2[B]$. Setting this to zero gives $k_1[A] = k_2[B]$, which flawlessly rearranges to $[B] \approx \frac{k_1}{k_2}[A]$.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes